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STD 11 - 6.2. Equilibrium - II (icon Equilibrium) — NEET STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceNEETSTD 11 - 6.2. Equilibrium - II (icon Equilibrium)4 Marks
MCQ
Solid $Ba(NO_3)_2$ is gradually dissolved in a $1.0 \times 10^{-4}\, M\, Na_2CO_3$ solution. At which concentration of $Ba^{2+}$, precipitate of $BaCO_3$ begins to form ? $(K_{sp}$ for $BaCO_3 = 5.1 \times 10^{-9})$
  • ✓
    $5.1 \times {10^{ - 5}}\,M$
  • B
    $7.1 \times {10^{ - 8}}\,M$
  • C
    $4.1 \times {10^{ - 5}}\,M$
  • D
    $8.1 \times {10^{ - 7}}\,M$
✓

Answer

Correct option: A.
$5.1 \times {10^{ - 5}}\,M$
a
Given $N{a_2}C{O_3} = 1.0 \times {10^{ - 4}}\,M$

$\therefore \,[CO_3^ - ] = 1.0 \times {10^{ - 4}}\,M$

$i.e.\,\,\,s = 1.0 \times {10^{ - 4}}\,M$

At equilibrium

$[B{a^{ +  + }}][CO_3^ - ] = {K_{sp}}\,of\,BaC{O_3}$

$[B{a^{ +  + }}] = \frac{{{K_{sp}}}}{{[CO_3^ - ]}} = \frac{{5.1 \times {{10}^{ - 9}}}}{{1.0 \times {{10}^{ - 4}}}}$

$ = 5.1 \times {10^{ - 5}}\,M$

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